TITLE: Proof Verification - Every sequence in $\Bbb R$ contains a monotone sub-sequence
QUESTION [25 upvotes]: Came across the following exercise in Bartle's Elements of Real Analysis. This is the solution I came up with. Would be grateful if someone could verify it for me and maybe suggest better/alternate solutions.
I also looked up these related questions - (1), (2), (3) - but was not happy with proofs given there. I seem to need some help understanding these. Any such help is appreciated. 

Show that every sequence in $\Bbb R$ either has a monotone increasing sub-sequence or a monotone decreasing sub-sequence.

Let $(x_n)$ be a sequence in $\Bbb R$. Suppose $(x_n)$ is not bounded. Without loss of generality we may assume that $(x_n)$ is not bounded above. Therefore given any real number there is a member of the sequence which is greater. Let $x_{n_1}$ be any member of the sequence. 
There is $x_{n_2} \gt \sup\{x_1, x_2, ..., x_{n_1} \}$. For $i \gt 1$ let $x_{n_i} = \{x_1, x_2, ..., x_{n_{i - 1}}\}$ then $(x_{n_k})$ forms a monotone subsequence of $(x_n)$. 
Now suppose instead that $(x_n)$ is bounded. By the Bolzano-Weierstrass Theorem there is a subsequence $(y_n)$ of $(x_n)$ which converges to a limit $y$. Without loss of generality there are infinitely many distinct values in $(y_n)$ that are unequal to $y$. Let $y_{k1}$ be the first such element. Let $y_{k2}$ be any element in $\{ y' \in (y_n) \ \ | \ \ |y' - y | \lt |y - y_{k1}|  \}$. 
For $i \gt 1$ let $y_{ki} \in \{ y' \in (y_n)  \ \ | \ \ |y' - y | \lt |y - y_{k \ i - 1}|  \}$. Such $y_{ki}$ exists for every $i \in  \Bbb N$ since $ \lim (y_n) = y $. Now let $(y_{kn})$ be the sub-sequence of $(y_n)$ thus formed. At least one of the two following sets must contain infinitely many elements. 

$\{ y \in (y_{kn}) \ \ | \ \ y \gt x\}$
$\{ y \in (y_{kn}) \ \ | \ \ y \lt x\}$

The one which does forms a monotone subsequence.

REPLY [3 votes]: I think the other proofs are superior, but this is a sketch of the method I came up with when trying to solve it on my own:
If $a_n$ is unbounded, we are done.
Suppose $a_n$ is bounded and let $R_N = \{a_n| n \ge N \}$  (i.e. the range of $a_n$ after $N$). Since $R_N$ is nonempty and bounded, it has a supremum. Note that for every $N$,  $\sup R_{N+1} \le \sup R_N$. Thus if $\forall N \sup R_N$ is a member of the sequence, then we can construct a decreasing subsequence. 
Now to finish up, suppose that for some $N$, $x = \sup R_N$ is not a member of the sequence. From the definition of supremum, we know that $\forall \epsilon >0$ there exists an element of the sequence, $a_n$, with $x- \epsilon <a_n$, and we can build an increasing subsequence based on that.